TS EAMCET · Maths · Circle

If the circles $C_{1} : x^{2} + y^{2} + 2 x + 4 y - 20 = 0, C_{2} : x^{2} + y^{2} + 6 x - 8 y + 9 = 0$ have $n$ common tangents and the length of the tangent drawn from the centre of similitude to the circle $C_{2}$ is $l$ then $\frac{l}{n^{2}} =$

A $4 \sqrt{39}$
B $\sqrt{39}$
C $\frac{\sqrt{39}}{4}$
D $2 \sqrt{39}$

Verified Solution

Answer & Solution

Correct Answer

(C) $\frac{\sqrt{39}}{4}$

Step-by-step Solution

Detailed explanation

Given, The circles C1:x2+y2+2x+4y-20=0 whose centre will be -1,-2 radius r1=12+22+20=5 and C2:x2+y2+6x-8y+9=0 whose centre is -3,4 and radius r2=9+16-9=4, Now we cans see C1C2=40, r1+r2=9 & r1-r2=1 So, r1-r2<C1C2<r1+r2 so there will be 2 common…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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If the circles C1:x2+y2+2x+4y-20=0, C2:x2+y2+6x-8y+9=0 have n common tangents and the length of the tangent drawn from the centre of similitude to the circle C2 is l then ln2=

Step-by-step Solution

If the circles $C_{1} : x^{2} + y^{2} + 2 x + 4 y - 20 = 0, C_{2} : x^{2} + y^{2} + 6 x - 8 y + 9 = 0$ have $n$ common tangents and the length of the tangent drawn from the centre of similitude to the circle $C_{2}$ is $l$ then $\frac{l}{n^{2}} =$